Dhaka Higher Secondary 2011/8
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Dhaka Higher Secondary 2011/8
$N$ represents a nine digit number each of whose digits are different and nonzero. The number formed by its leftmost three digits is divisible by $3$ and the number formed by its leftmost six digits is divisible by $6$. It is found that $N$ can have $2^k3^l$ different values. Find the value of $k + l$.
 Tahmid Hasan
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 Location: Khulna,Bangladesh.
Re: Dhaka Higher Secondary 2011/8
plz clear why a+b+c has $3^5$ choice?
A man is not finished when he's defeated, he's finished when he quits.
 Tahmid Hasan
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 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Dhaka Higher Secondary 2011/8
i got the ans wrong
didn't notice the condition that all the numbers were distict.
didn't notice the condition that all the numbers were distict.
বড় ভালবাসি তোমায়,মা

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 Joined: Thu Jan 20, 2011 10:46 am
Re: Dhaka Higher Secondary 2011/8
You have forgotten about last 3 differnts digit.They have also 6 permutation.Then you will get the correct ans k+l=12
Re: Dhaka Higher Secondary 2011/8
Let me show how I advanced..... 1st I divided the numbers into 3 groups using mod i.e 0 mod 3, 1mod 3 and 2 mod 3. According to divisibility rule, numbers made up by 1st 3 digits/3 & numbers made up by 2nd 3 digits/3 & 2 both. so, 6th digit must be even. 2/4/6 or 8. for 2 mod 3 digits for the 2nd group,we can have 2x2! permutations and so we have 2x3! permutations for the leftmost 3 digits... for 0 mod 3 & 1 mod 3 digits for the 2nd group we can have 2! permutations each for the 2nd group and 2x3! permutations for the leftmost 3 digits...
Now let's consider that the 2nd group of digits be filled up with 0,1 & 2 mod 3 digits each. so for 2 mod 3 digits we have 2 x 3C1 x 3C1 x 2! permutations and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st three digits.
for 0 mod 3 and 1 mod 3 we have 3C1 x 3C1 x 2! permutations for the 2nd group and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st group respectively.....
multiplying the related permutations and adding them we have
4x12 + 36x48 + 2x12 + 18x48 + 2x12 + 18x48
= 2^5 x 3 x 37!!!
(That's the problem with my solution... Here comes a prime number!!! I'm kinda sure that my method's right. Can anyone help me in this case? Is my solution wrong or is the ques wrong???? )
Now let's consider that the 2nd group of digits be filled up with 0,1 & 2 mod 3 digits each. so for 2 mod 3 digits we have 2 x 3C1 x 3C1 x 2! permutations and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st three digits.
for 0 mod 3 and 1 mod 3 we have 3C1 x 3C1 x 2! permutations for the 2nd group and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st group respectively.....
multiplying the related permutations and adding them we have
4x12 + 36x48 + 2x12 + 18x48 + 2x12 + 18x48
= 2^5 x 3 x 37!!!
(That's the problem with my solution... Here comes a prime number!!! I'm kinda sure that my method's right. Can anyone help me in this case? Is my solution wrong or is the ques wrong???? )
Re: Dhaka Higher Secondary 2011/8
Let me show how I advanced..... 1st I divided the numbers into 3 groups using mod i.e 0 mod 3, 1mod 3 and 2 mod 3. According to divisibility rule, numbers made up by 1st 3 digits/3 & numbers made up by 2nd 3 digits/3 & 2 both. so, 6th digit must be even. 2/4/6 or 8. for 2 mod 3 digits for the 2nd group,we can have 2x2! permutations and so we have 2x3! permutations for the leftmost 3 digits... for 0 mod 3 & 1 mod 3 digits for the 2nd group we can have 2! permutations each for the 2nd group and 2x3! permutations for the leftmost 3 digits...
Now let's consider that the 2nd group of digits be filled up with 0,1 & 2 mod 3 digits each. so for 2 mod 3 digits we have 2 x 3C1 x 3C1 x 2! permutations and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st three digits.
for 0 mod 3 and 1 mod 3 we have 3C1 x 3C1 x 2! permutations for the 2nd group and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st group respectively.....
multiplying the related permutations and adding them we have
4x12 + 36x48 + 2x12 + 18x48 + 2x12 + 18x48
= 2^5 x 3 x 37!!!
(That's the problem with my solution... Here comes a prime number!!! I'm kinda sure that my method's right. Can anyone help me in this case? Is my solution wrong or is the ques wrong???? )
Now let's consider that the 2nd group of digits be filled up with 0,1 & 2 mod 3 digits each. so for 2 mod 3 digits we have 2 x 3C1 x 3C1 x 2! permutations and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st three digits.
for 0 mod 3 and 1 mod 3 we have 3C1 x 3C1 x 2! permutations for the 2nd group and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st group respectively.....
multiplying the related permutations and adding them we have
4x12 + 36x48 + 2x12 + 18x48 + 2x12 + 18x48
= 2^5 x 3 x 37!!!
(That's the problem with my solution... Here comes a prime number!!! I'm kinda sure that my method's right. Can anyone help me in this case? Is my solution wrong or is the ques wrong???? )
 bristy1588
 Posts: 92
 Joined: Sun Jun 19, 2011 10:31 am
Re: Dhaka Higher Secondary 2011/8
${24*(2^5*3^3+2^3*3)}$
I get this as the answer, What is the actual answer>??
I get this as the answer, What is the actual answer>??
Bristy Sikder
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Re: Dhaka Higher Secondary 2011/8
my answer is also $10$
i have gone case by case.
i have gone case by case.
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Dhaka Higher Secondary 2011/8
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